Organic Chemistry Acids and Bases #2

Which of the following types of reactions will be favored by t-butoxide reacting with a secondary halide? 

A. S_N2          

B. E2          

C. E1          

D. S_N1

Answer:  B

SHORT-AND-SWEET:

When it comes to the substitution and elimination reactions, which particular reaction will occur is largely determined by the attacking nucleophile/base, so always start there!  Our attacker is t-butoxide, a very bulky molecule, whose negative charge makes it a strong attacker.  But will it be a strong nucleophile or a strong base?  Most of the time, the basicity and nucleophilicity track together.  Rarely, however, this is not the situation.  T-butoxide is the perfect example!  Because it is so bulky, it will be very hard for it to get to the heart of the electrophile, and attack the carbon -- therefore, it is a weak nucleophile.  However, it will have a very easy time stealing a proton, which makes it a strong base.  

In general, a strong attacker will favor a single-step reaction (either SN2 or E2) -- a strong base will promote E2, and a strong nucleophile will promote SN2.  Because t-butoxide is such a strong base it will promote E2 (answer B).

If the attacker is not this strong, the other factors to consider are the bulkiness of the substrate (only SN2 will prefer a less bulky substrate), and the solvent.

THE WHOLE STORY:

The substitution and elimination reactions are reactions between nucleophiles and electrophiles.  Nucleophiles (Lewis bases) are the electron-rich molecules, and they love the positively charged nucleus, to which they like to donate their lone pair of electrons.  Electrophiles (Lewis acids) accept these electrons because, like their name suggests, they love electrons!

The best analogy for reactions between nucleophiles and electrophiles is between guys and girls.  The "attacker" is typically the guy -- let's call him Justin.  He is our nucleophile, and he approaches the electrophile, the girl -- we'll call her Jane.  Let's see what happens with Justin and Jane.....

In SUBSTITUTION, the end result of nucleophilic attack is that the nucleophile substitutes for a group on the electrophile, i.e. Justin substituting for Jane's boyfriend!  

•  In SN2, the nucleophile attacks the electrophile, at the same time kicking the leaving group out.  In this dramatic scenario, Justin walks in, and the boyfriend gets the boot, all at the same time! 
• In SN1, the leaving group leaves first, and the remaining carbocation is attacked by the nucleophile.  Here, Jane's boyfriend leaves first, and then Justin swoops in!

In ELIMINATION, the end result of the attack is the elimination of a group on the electrophile, with the formation of a new double bond.  In the Justin-Jane story, elimination would lead to the boot to the boyfriend, and Jane deciding to be single for some time! 

• In E1, the leaving group leaves first, leaving a carbocation.  The base comes next, stealing a proton from the carbon adjacent to the positively charged carbon, facilitating the double bond.  In our love triangle that translates into the boyfriend deciding to leave Jane, followed by Justin coming by and snatching....not Jane's heart, but Jane's friend, the proton!
• In E2, the base steals the proton and the leaving group leaves at the same time.  In this scenario, Jane's proton gets stolen by Justin, and her boyfriend leaves her.  That's what we call a bad day!

Why are there two elimination reactions and two substitution reactions?  The answer has to do with the kinetics!  SN2 and E2 have the 2nd-order kinetics, which means that the reaction rate depends both on the nucleophile/base, and the electrophile substrate.  Why?  Because these reactions occur in a single step, which has a high-energy transition state.  This "transition" molecule in which the bond between the nucleophile and electrophile is forming, and the bonds within the electrophile are breaking requires presence of both reactants at the same time.

On the other hand, SN1 and E1 have the 1st-order kinetics, and their reaction rate depends on the substrate only.  These reactions occur in two steps, the first of which involves only the substrate.  Only after the substrate had finished the first step does the nucleophile/base step in.

How can we determine which reaction will dominate -- substitution or elimination, first-order or second-order?  Often, it's enough to look at the ATTACKER (NUCLEOPHILE / BASE), so always start there!

Because the reaction rate in SN1 and E1 depends only on the substrate/electrophile, how good the attacking nucleophile is does not matter.  On the contrary, SN2 and E2 will happen only with a strong nucleophile/ base around.  Why?  If you have a strong nucleophile/base, it will not sit around waiting for the substrate to do the first step on its own.  A weak nucleophile/base will be perfectly content with twiddling its thumbs while the substrate completes the first step independently.  If our friend Justin is super cool and confident, he won't wait around for Jane to dump her boyfriend (or get dumped).  Oh, no, no, no!  He will step in and do what he's gotta do!  On the other hand, if Justin is not as confident, he'll be OK with letting Jane get rid of the boyfriend, and then he will come in.

Hold on for a minute!  We keep talking about Justin acting as a nucleophile and a base.  "The two are the same, right?", you ask.  Though they are frequently thought of as the same thing, they are not!  The difference between a nucleophile and a base is in what each wants: a base wants to bind a proton, and a nucleophile wants to bind a carbon atom.  Because a base comes just close enough to the molecule to snatch a proton (which usually hangs out peripherally), the bulkiness of a molecule will not affect how strong a base it is.  On the other hand, a nucleophile needs to attack the carbon directly, and if it has a lot of side chains, it can forget about it!  A way that we can tie this to our analogy is that Justin The Nucleophile is interested in getting to Jane.  Justin The Base is interested in getting anything -- he will be satisfied with the proton sitting on the side.

Because of this, the reactions preferred by a strong nucleophile will differ from reactions preferred by a strong base.  Because a strong base will be satisfied with snatching a proton and going about its business, it will prefer elimination E2.  A strong nucleophile will want to attack a carbon, so it will promote substitution SN2.

Remember:  if you have a STRONG nucleophile or a STRONG base, the reaction will go with the second-order kinetics (number 2 = SN2 or E2, respectively).

After analyzing the attacker, what should you look at next?  The SUBSTRATE -- its overall structure and its leaving group!

Side chains on the substrate/electrophile are like Jane's girlfriends -- they intimidate Justin!  Similarly, a bulky substrate, like a tertiary substrate, will not allow enough room for a nucleophile to approach and attack in the SN2 fashion.  A methyl or a primary substrate, on the other hand, will be very inviting for a nucleophilic attack.

A bulky substrate will be perfect for the first-order kinetics reactions, such as SN1 and E1, because these reaction involve carbocation formation.  A tertiary substrate, with multiple carbon side chains, will help pull away some of the positive charge, and will stabilize the carbocation.  And the more stable the intermediate, the more likely will the reaction occur.  Therefore, tertiary substrate (and resonance stabilization) are ideal for SN1 and E1.  It's like breaking up with a boyfriend or a girlfriend -- after that happens, it's nice to have friends around to cheer you up, right?

The tertiary substrate will work best for E2 as well, because it the most substituted alkene possible is preferred (Zaitsev rule), which is easiest to do with a tertiary substrate.

All four reactions require a good leaving group, a molecule that will be stable on its own when it leaves (read: a mentally stable person!  Otherwise, the boyfriend will cling on to Jane indefinitely!).  A good leaving group will be a weak base (a conjugate base of a strong acid).  Halides are great leaving groups, unlike -OH, -NH2, or alkoxides (RO-), which are terrible!

The last factor is the solvent in which the reaction occurs.  A solvent affects nucleophile strength.  Let's go back to Justin for a second.  Justin (whose last name is Bieber -- we forgot to mention, oops!), has his own girl fan base surrounding him at all times.  If he is attracted to these girls, what will happen with his interest in approaching Jane?  It will dissipate.  What if he is not so attracted to these other girls?  Well, that will help him be more determinate about approaching Jane.  Let’s look at an example of a reaction.  In a protic solvent (one that has -OH or -NH2 groups) SN2 is less likely to happen, because protic molecules create a cage around the nucleophile (attractive girls around Justin), making it less interested in attacking another molecule.  Therefore, aprotic solvent is perfect for SN2!  SN1, on the other hand, will be promoted by a protic solvent, which will stabilize its carbocation.


Let’s finally go back to our question, in which t-butoxide is attacking a secondary halide!  Where did we say you should start?  With the attacker!  Look at t-butoxide!  

t-butoxide is a negatively charged molecule, so it will want to bind something positive, either a proton or a carbon.  The former makes it a strong base, and a STRONG base  will favor the second-order kinetics, either SN2 or E2 (eliminate C and D).  But which one?  As a strong base it could undergo E2 easily.  What about SN2?  Because t-butoxide is very bulky – just look at those three methyl groups – it will not be able to squeeze in all the way to the carbon of the secondary halide, so there will be no substitution (eliminate A)!  Which is why it will sweep by, steal the proton, which will facilitate the formation of the double bond, and boot the leaving group out!  E2 is the correct answer (answer B).

THE BIG PICTURE:

1.  Difference between a nucleophile and a base --  a nucleophile wants to bind a carbon, and the base would rather bind a proton.  The former will promote substitution, and the latter will promote elimination.

2.  Follow the algorithm:  first decide between number 2 (SN2, E2) and number 1 (SN1, E1) by looking at the strength of the attacker.  A strong attacker (either a nucleophile or a base) will want to attack immediately, and will want to finish the reaction quickly, in a single step (NUMBER 2).  The opposite will happen with a weak attacker (number 1).  The next decision is substitution versus elimination.  A bulky attacker will prefer elimination. 


3.  The molecular structure tells the story!  Bulkiness means a lot of steric hindrance, not much room to work in (elimination sounds good in this scenario), but it also stabilizes your carbocation (SN1 works, also).

~The MCATPOD Team~

Electrochemistry #2

Based on the information provided in the table below, which of the following reactions can be used to build an electrolytic cell?

A. Ni(s) + 2Ce⁴⁺(aq) → Ni²⁺(aq) + 2Ce³⁺(aq)

B. Cr₂O₇^2-(aq) + 14H⁺(aq) + 6I^-(aq) → 2 Cr³⁺(aq) + 3I₂(s) + 7H₂O(l)

C.  Fe²⁺(aq) + Cu(s) → Fe(s) + Cu²⁺(aq)

D. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Half-reaction	Standard potential
Ce3+(aq) → Ce4+(aq) + e-	-1.61V
Fe2+(aq) + 2e- → Fe(s)	-0.44V
Cu(s) → Cu2+(aq) + 2e-	-0.34V
Ni2+(aq) + 2e- → Ni(s)	-0.28V
I2(s) + 2e- → 2I-(aq)	+0.54V
Zn(s) → Zn2+(aq) + 2e-	+0.76V
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H20(l)	+1.33V

Answer: C.

SHORT-AND-SWEET:

The idea behind the electrolytic cell is using energy from an external voltage source to force a non-spontaneous redox reaction.  So, in order to answer the question we need to figure out which of the four offered reactions is non-spontaneous.  Because it is non-spontaneous, this reaction will have a negative electromotive force (emf) or potential E⁰.  Cell potential E⁰ we can calculate from the half-cell potentials:

E⁰ = E⁰_{red (reduction half-reaction)} - E⁰_{red (oxidation half-reaction)}   

Let's look at our answer choices.  In A, the reduction half-reaction is Ce⁴⁺ reduction to Ce³⁺, and the oxidation half-reaction is Ni oxidation to Ni²⁺.  The table suggests that the standard oxidation potential of Ce⁴⁺ is -1.61V.  Simply change the sign to get the E⁰_red: +1.61V.  The E⁰_red of Ni²⁺ is -0.28V, so the cell potential E⁰ = 1.61V - (-0.28V) = +1.89V, which means a spontaneous redox reaction.  Ideal for a galvanic cell, but not for electrolytic cell (eliminate A).  We can use the same process with the reaction in the answer choice B, whose cell potential is +0.79V, and answer choice D, whose cell potential is +1.10V, which eliminate both.  In the answer choice C, Fe²⁺ is reduced, and its E⁰_red is -0.44V.  Cu is oxidized, and the E⁰_red of this half-reaction is +0.34V, giving E⁰ = -0.44V - 0.34V = -0.78V.  Negative E⁰ corresponds to a non-spontaneous reaction, which could be used to build an electrolytic cell (answer C).
Things to keep in mind:

• When calculating the standard cell potential, always use standard REDUCTION potential for each half-reactions!  If you are given a table, make sure you note whether you are given reduction or oxidation potentials (or a mix of both).  If you are given the E⁰_ox, just flip the sign to get the E⁰_red!
• E⁰_red is an intensive property, which means that the stoichiometric coefficient in a half-reaction does not affect its value!

THE WHOLE STORY:

Electrochemistry deals with the reduction-oxidation (redox) reactions, which involve transfer of electrons. 
Oxidation occurs when a compound gives away an electron.  The compound that receives this electron is reduced (reduction).

REMEMBER: Oxidation and reduction cannot occur independently of one another!!!  An electron has to come off of one compound (oxidation), and it has to be picked up by another (reduction).

The chemical compounds that participate in this reaction have the appropriate names, based on what they do to the substance they are reacting with.

• Oxidizing agent (oxidant) → it oxidizes the other substance by taking electrons away from it.  By taking these electrons, the oxidizing agent gets reduced.
• Reducing agent (reductant) → it reduces the other substance by giving away its electrons.  By doing this, the reducing agent gets oxidized.

How can we tell whether a particular substance will be an oxidizing or reducing agent, whether it will be reduced or oxidizes?  By knowing what other substance it has to react with!  

The following analogy can help illustrate this.  Let's say that you have two neighboring countries (our chemical compounds), and one of them wants to occupy a part of the other.  The part of the land that might get occupied is our electron, and the country that wants to occupy it is -- you guessed it -- the oxidizing agent!  Now, what will determine whether the electron (or the land) will get occupied?  Two things: how strongly the attacking country wants it, and how willing the defending country is to defend it. 
If the attacker really wants this land, it acts like a good oxidizing agent, which really wants to be reduced! 
If the defending country doesn't particularly love this piece of land (and they want to get rid of it -- we know, unlikely to happen in the real world, but let's pretend it's possible!), it acts like a good reducing agent, which really wants to get rid of its electron and be oxidized!      

Standard reduction potential E⁰_red tells us how badly a substance wants to be reduced -- compared to the standard hydrogen electrode.  The more positive the E⁰_red, more the substance wants to be reduced.  Conversely, the more negative the E⁰_red, less interested the substance is in being reduced.

Standard reduction potential is the one that you will most commonly hear about.  However, there is also its counterpart -- the standard oxidation potential (E⁰_ox), which tells you how badly a substance wants to be oxidized.  A substance that really wants to be oxidized will have a very positive E⁰_ox. Be very careful about noting whether the standard potential refers to reduction or oxidation.  Though this measure is usually expressed as the standard reduction potential (e.g. Hg²⁺ + 2e^- → Hg(l)  E⁰ = +0.85V), sometimes MCAT will try to trick you on this (like we did in our question!), by expressing it as Hg(l) → Hg²⁺ + 2e^-    E⁰ = -0.85V.  Mercury is clearly being oxidized here, so the standard potential is that of oxidation.
 

The tables of standard potentials can tell you how particular substances will react with one another, if left to their own devices.  It's like the analogy above -- the land (or electron) in question will end up in the hands (or orbitals) of the country (or compound) wanting it more!  Let’s look at an example of a reaction between platinum and copper.  E⁰_red of Pt²⁺ + 2e^- → Pt is +1.19V, whereas E⁰_red of Cu²⁺ + 2e^- → Cu is +0.34V.  Clearly, the platinum cation wants to be reduced more, which is what will happen if platinum and copper react together:  Pt²⁺ + Cu(s) → Pt(s) + Cu²⁺.

Now that we understand the redox reactions, let's move on to the galvanic (voltaic) and electrolytic cells.
Make sure you understand the fundamental difference between the two:

• Galvanic cell uses a spontaneous redox reaction, and transforms the chemical energy of the reaction into electrical energy, which enables it to do work.  Think: battery!
• Electrolytic cell requires energy to be put into the electrolytic cell to cause otherwise nonspontaneous redox reaction to occur.  

Here are some other useful things to know about these cells. 

• Regardless of the type of cell, oxidation always occurs on the anode, and reduction always occurs on the cathode (remember: consonants (R-C) go to one side, vowels (O-A) to the other!). 
• In the galvanic cell the electrons are released during oxidation on the anode and they flow through the external circuit to the cathode, which is why the anode is labeled as negative, and the cathode as positive.  We can envision electrons being attracted to the positive cathode from the negative anode!  (This, however, should not be interpreted as the charges on the electrodes -- the actual charges on the electrodes are essentially zero!)
• In the electrolytic cell the voltage source's negative terminal is connected to the cathode, which from there receives electrons to reduce a substance.  The electrons that are removed during oxidation at the anode travel to the positive terminal of the voltage source, completing the circuit.  Therefore, cathode is labeled as negative, and anode as positive.  

Now, back to our question!  You are asked to identify a reaction which could be used to build an electrolytic cell.  As we said above, for electrolytic cell one needs a nonspontaneous reaction.  Non-spontaneous reaction corresponds to a cell whose cell potential E⁰ is negative.  Cell potential refers to the difference between the potentials of the half-cells:  

E⁰ = E⁰_{red (reduction half-reaction)} - E⁰_{red (oxidation half-reaction)}  

What does this mean in practice? Let's look at the answer choice A.  The reduction half-reaction is Ce⁴⁺ reduction to Ce³⁺, and the oxidation half-reaction is Ni oxidation to Ni²⁺.  From the table we see that the standard oxidation potential of Ce⁴⁺ is -1.61V.  In order to get the E⁰_red, we just flip the sign, making it +1.61V.  The E⁰_red of Ni²⁺ is -0.28V, giving the E⁰ of : +1.61V - (-0.28V) = +1.89V, which means a spontaneous redox reaction.  Ideal for a galvanic cell, but not for electrolytic cell (eliminate A)!  

One thing to note here.  In this reaction the stoichiometric coefficient in front of Ce⁴⁺ and Ce³⁺ is 2.  Commonly folks make the mistake of multiplying the E⁰_red of a half-reaction by the corresponding stoichiometric coefficient.  E⁰_red is an intensive property, which means that the stoichiometric coefficient in a half-reaction does not affect its value.

We can use the same process with the reaction in the answer choice B, whose cell potential is +0.79V, and answer choice D, whose cell potential is +1.10V (eliminate B and D).  Now, let's look at the answer choice C.  Fe²⁺ is reduced, and its E⁰_red is -0.44V.  Cu is oxidized, and the E⁰_red of this half-reaction is +0.34V.  Cell potential E⁰ will be -0.44V - 0.34V = -0.78V, which means that this is a non-spontaneous reaction, which could be used to build an electrolytic cell (answer C).

  
BIG PICTURE: 

1. Galvanic cell = spontaneous reaction = does work!  Electrolytic cell = non-spontaneous reaction = requires energy from another source!

2. When using standard potential to calculate cell potential E⁰, always use standard REDUCTION potential!  If you are given standard oxidation potential, just flip the sign to get the reduction potential.

3.  Stoichiometric coefficients in a half-reaction will not change the value of E⁰_red!
 
~The MCATPOD Team ~

Female Reproductive System #2

All of these statements about female gametogenesis are correct, EXCEPT:

A.  Just like in the case of the spermatogonia, the oogonia population is renewed throughout adult life.
B.  Primary oocyte is a diploid cell.
C.  Primary oocyte remains in prophase I until ovulation.
D.  Secondary oocyte comes out of metaphase II only if fertilization occurs.

Answer: A.

SHORT-AND-SWEET:

Female gametogenesis beings during the fetal life, with ovarian germ cells dividing mitotically to produce oogonia.  However, while the female fetus is still in utero, the germ cells will produce a finite number of oogonia and then stop the production (unlike spermatogonia production, which continues throughout the adult life -- answer A).  Diploid oogonia will then undergo meiosis, and arrest in prophase I as diploid primary oocyte.  Ovulation will trigger primary oocyte to continue through meiosis, but it will arrest again in metaphase II as the haploid secondary oocyte, to continue development into haploid mature ovum only if fertilization occurs.  

THE WHOLE STORY:

Before reviewing female gametogenesis (oogenesis), let's make sure we have solid understanding of the more fundamental matters.

First, let's clarify the cellular DNA content.

• With the exception of gametes, all other healthy human cells have 23 pairs of chromosomes -- each pair consisting of one Mom chromosome and one Dad chromosome, giving a grand total of 46 chromosomes per cell.  These cells are considered DIPLOID (2n).
• When the DNA replicates, each of the parent's chromosomes will consist of two identical sister chromatids.  That means that the cell which has undergone DNA replication will still have 46 different chromosomes, but each chromosome will consist of two sister chromatids (92 chromatids total per cell).  However, this cell is still considered DIPLOID (2n), because it has 46 chromosomes total.
• HAPLOID (n) cells, such as gametes, will contain only one copy of each chromosome (essentially, 23 chromatids).

Moving on to discuss M&M -- mitosis and meiosis.  These are the two types of cellular division.  Both are preceded by one round of DNA replication.

• Mitosis consists of one round of cell division following DNA replication, and it produces two daughter cells which are identical to their mother cell (all diploid, with 23 pairs of chromosomes).  Most of the cells in the human body which can replicate do so by mitosis.
• Meiosis, which occurs only in reproductive cells, consists of two sequential rounds of cell division (meiosis I and meiosis II) following DNA replication.  The final result of meiosis are four haploid gametes.  In male, these are four sperm cells, whereas female produces one ovum and three polar bodies.  During meiosis I there will be an opportunity for exchange of Mom's and Dad's genetic material, which ensures genetic diversity. 

Now that we understand the basics, let's look at what happens during oogenesis.  
Female gametogenesis begins during the fetal period.  The ovarian germ cells divide mitotically, producing multiple identical diploid oogonia.  Once they have produced several million oogonia, they stop dividing -- and ALL BEFORE BIRTH!  This is one of the key differences from the male gametogenesis, where the germ cells continue producing spermatogonia throughout adult life.  This helps us pick out answer choice A as the correct answer.  (If you were on the actual test you would go on to the next question......  How nice.  Wish every other question was this easy!)

But let's review the rest of the oogenesis.  In order to produce its progeny, oogonium cell will enter meiosis -- meiosis I, to be more specific.  However, it will not get far.  It will arrest in the first phase of meiosis I, prophase I, as the PRIMARY OOCYTE.

How "much" DNA does the primary oocyte have?  Well, we know that the DNA was replicated prior to meiosis, so the cell will have total of 46 chromosomes (92 sister chromatids).  Such cell is still considered diploid.  There is no division of genetic material during prophase I, which means that primary oocyte will also be diploid (eliminate answer choice B).

The lazy primary oocyte will remain in prophase I for years, until puberty and the first ovulation (eliminate answer choice C).  Just before ovulation the "selected" primary follicle and its primary oocyte will complete meiosis I, producing one SECONDARY OOCYTE (and one polar body to be discarded), which heads into meiosis II.  

What about secondary oocyte's DNA content?  What transpires during the rest of meiosis I?  The DNA is first exchanged between the homologous Mom and Dad chromosomes, and then during anaphase I, the homologous chromosomes (each consisting of two chromatids) are separated, yielding two haploid daughter cells, each with 23 chromosomes (either of Mom or Dad origin, each having two chromatids).

The secondary oocyte likes to chill out too, and arrests not long after its formation -- in metaphase II (good bye, answer choice D), and it stays there until the ovulated egg is fertilized by sperm.  While the sperm is trying to break in, the haploid secondary oocyte hurries to finish meiosis II, producing haploid ootid (and another polar body), which has one copy of each chromosome.

The sperm and ootid nuclei will not fuse until the second polar body is released, and the ootid matures producing a mature ovum.  After their nuclei fuse, they will produce a diploid fertilized egg, or a zygote.

Basically, you can think of oogenesis as one big arrested development (like the TV show!), with lazy primary and secondary oocytes napping whenever they can.  But how to remember when each will take a nap?  Easy -- number 1 and number 2!
Number 1:  The PRIMARY oocyte will arrest in the FIRST phase (prophase) of meiosis I.
Number 2:  SECONDARY oocyte will arrest in the SECOND phase (metaphase) of meiosis II.

Be selected or die is the other motto of oogenesis!  Only the primary oocyte which gets "selected" for ovulation in a particular month continues its development.  All of the other ones prepping for that month's ovulation die.  Similarly, the ovulated oocyte arrested in metaphase II needs to be fertilized in order to continue its development.  If there is no fertilization, it will die a sad and lonely death.

BIG PICTURE:
1.  Diploid cells have 23 pairs of chromosomes, each pair consisting of one Mom and one Dad chromosome.  Haploid cells have 23 chromosomes, each either being of Mom origin, Dad origin, or mixed Mam/Dad origin (due to recombination which occurs during meiosis).

2.  DNA replication precedes both mitosis and meiosis.  Mitosis consists of one round of cell division, which produces two daughter cells, genetically identical to their mother cell (all 2n).  Meiosis consists of two rounds of cell division, producing four haploid cells.

3.  Oogenesis is one big arrested development show!  Diploid oogonium produces diploid primary oocyte, which arrests in prophase I.  Ovulation pushes it forward, but the oocyte arrests again in metaphase II as the secondary oocyte.  Finally, if it gets fertilized it will continue its development to mature ovum.

~The MCAT POD Team~

Fluids #2

A pipe carries water from a hot-water tank in the basement to the three floors above it.  Assume that both the cross-sectional area of the pipe and the water flow rate are constant.  Which of the following statements will be true about water exiting the pipe on the top floor of the building?

A.  The speed of the hot water will be higher than at the basement tank level.
B.  The speed of the hot water will be lower than at the basement tank level.
C.  The pressure of the hot water will be higher than at the basement tank level.
D.  The pressure of the hot water will be lower than at the basement tank level.

Answer: D

SHORT-AND-SWEET:

When you see a question that talks about different types of energy in a system, think of conservation of energy!  When applied to the flowing fluids, this principle is called the Bernoulli's principle.  The principle states that the sum of all the different kinds of energies of the flowing fluid will be constant.  What are the types of energy a flowing fluid can have?  Kinetic (because it is moving), potential (because it might be moving from one height level to another), and "pressure energy" (energy expended by fluid to exert pressure on the container walls).  Because their sum is always constant, an increase in one type of energy will come at the expense (decrease) of the other types of energy.

In the question above, water's potential energy will increase (because the water is flowing from the basement to the top floor) -- at the expense of either water's kinetic energy or pressure.  We are explicitly told that the flow rate (Q) and the cross sectional area (A) are constant, which means that the water velocity will remain constant as well (recall the expression for the flow rate: Q = A v).  That also means that there will be no change in kinetic energy.  Therefore, increase in the potential energy will have to come at the expense of the water pressure, leading to water pressure drop on the higher floors (answer D).

THE WHOLE STORY:

This week we continue to examine fluids.  More specifically, we are dealing with hydrodynamics: fluids in motion.  Like we said last time, when you hear the word "dynamics" think about kinetic energy.  But that's not all we have in this question.  Water moving from basement level to the top floor suggests some height change, which makes you think of gravitational potential energy.  It also talks about pressure.  Of course, you have to talk about pressure when you talk about fluids.

Though not explicitly stated, the question treats the building-water tank-pipe-water system as a closed system, in which there will be conservation of energy.  When we talk about conservation of energy for fluids, we are talking about Bernoulli's principle.

Bernoulli's principle applies to ideal fluids.  Let's recall characteristics of an ideal fluid:
- steady flow = all fluid molecules are moving at the same speed.
- incompressible = under compression, it will not change volume (due to repulsive forces between molecules).
- non-viscous flow = all molecules move linearly in laminar flow.

In the MCAT, unless explicitly told otherwise, assume that all fluids you encounter are ideal!

Back to good ol' Bernoulli and conservation of energy in flowing fluids.  What are the different kinds of energy that a flowing ideal fluid can have?  Well, for starters, given the "flowing" part, it can definitely have KINETIC ENERGY.  What else?  Think of a river flowing from a mountain spring towards an ocean.  It is flowing from a higher elevation to a lower ground, which points to it POTENTIAL ENERGY.  What else?  When we talk about fluids, we often talk about their pressure, which is essentially a measure of how hard the fluid molecules are pushing against the walls of the fluid container.  So, the fluid has to have some energy spared to exert pressure, so let's call it exactly that: "ENERGY to exert PRESSURE".

The conservation of energy principle says that in a closed system total energy will be constant.  Therefore, the sum of all the different kinds of energies will be constant as well.  For a flowing fluid that means:

ENERGY TO EXERT PRESSURE + KINETIC ENERGY + POTENTIAL ENERGY = CONSTANT

So, if the fluid flows faster and therefore expends more kinetic energy, it will have less energy to exert pressure on the walls of its container, and the pressure will drop (if the height does not change).

Let's use a more colorful example, that of a hamster (we'll name him Bernoulli) in a cage.
Bernoulli can have fun in his cage in several different ways.  He can run in the cage.  He can climb the walls of the cage (a pretty athletic hamster, huh?), or he can push against the walls of the cage (trying to escape, perhaps?).  Now, if Bernoulli wants to be running as fast as he can (kinetic energy), he won't have much energy left to be climbing the walls of his cage (gravitational potential energy).  If he decides to push, push, push against the walls of the cage, it will be challenging for him to be trying to run like a wind at the same time.

Now, let's look at the actual Bernoulli's equation:

The first element in the equation, P, is clearly the pressure of the fluid.  The second one looks familiar, too, doesn't it?  If you substituted the fluid density ρ for mass m, this part would look just like the kinetic energy expression that you had seen a million times.  The same thing with the last element in the equation, which looks much like the expression for potential energy (when you substitute ρ for m).

Now, let's go back to our question.  We know that in this system in which the water is moving from the basement tank to the top floor the total energy will be conserved.  So, whatever energy will be expended in the system to get the water to the top floor (potential energy) will come at the expense of either kinetic energy or the "pressure energy".  But which one will have to give?

Note that in the question itself you are told that neither the flow rate nor the cross-sectional area of the pipe change.  OK, so it's the same pipe, with the same cross-sectional area.  But what does the constant flow rate mean?  Flow rate tells us the amount of fluid flowing through a pipe in one second.  Constant flow rate means that whatever fluid volume enters the pipe in one second, the same volume of fluid will come out at the other end in that second.  The continuity equation also holds true in this case (1 and 2 refer to different points along the pipe, for example the basement and the top floor):

A1 v1 = A2 v2

If the cross-sectional area does not change (A1 = A2), it also means that velocity along the pipe will not change either (v1 = v2), which eliminates answers A and B.

However, it also means that fluid kinetic energy remains constant along the length of the pipe.  Therefore, the increase in the gravitational potential energy will come at the expense of the "pressure energy", and the water pressure on the top floor will be lower than in the basement tank (answer D).

BIG PICTURE:

1. When you see a question that talks about different kinds of energy, think conservation of energy (unless explicitly stated otherwise)!

2. Bernoulli's principle is the principle of conservation of energy applied to moving fluids.  Increase in one type of energy will come at the expense (decrease) of the other types of energy.  

~The MCAT POD Team~