Question

Write the name of the basic building units of proteins and nucleic acids. How can you differentiate between fibrous and globular proteins on the basis of their structures ?

Answer 1

Concept: This is a 'detective' question. We use each clue to nail down a group, and then put the pieces together. The key first clue is that a Grignard reagent is reacting with $\mathrm{C_3H_5N}$. A Grignard reagent adds to a nitrile, and after hydrolysis the nitrile turns into a ketone.

Step 1: Find compound (A)
The formula $\mathrm{C_3H_5N}$ has the right shape for a nitrile, and reacting with a Grignard reagent confirms it. So (A) is propanenitrile, $\mathrm{C_2H_5CN}$.

Step 2: Find compound (B)
Now read its tests carefully. (B) gives an orange-red precipitate with 2,4-DNP, so it has a carbonyl group ($\mathrm{C=O}$). It does not reduce Tollens' or Fehling's reagent, so it is not an aldehyde, it must be a ketone. It does not give the iodoform test, so it is not a methyl ketone (no $\mathrm{CH_3CO\text{-}}$ group). It does not decolourise bromine water, so there is no carbon to carbon double bond. Putting this together with the fact that a phenyl group came from $\mathrm{C_6H_5MgBr}$, (B) is propiophenone, $\mathrm{C_6H_5COC_2H_5}$.

Step 3: Find compound (C)
On drastic (strong) oxidation, the side chain attached to the benzene ring is chopped off and the ring carbon becomes $\mathrm{-COOH}$. This gives benzoic acid, $\mathrm{C_6H_5COOH}$, whose formula is $\mathrm{C_7H_6O_2}$, exactly as stated. So (C) is benzoic acid.

Step 4: Write the reaction of (A) with the Grignard reagent
The phenyl group of the Grignard adds across the $\mathrm{C{\equiv}N}$, and on hydrolysis we get the ketone plus ammonia: \[ C_2H_5C{\equiv}N \xrightarrow{C_6H_5MgBr} C_2H_5C(=N{-}MgBr)C_6H_5 \xrightarrow{H_3O^+} C_2H_5COC_6H_5 + NH_3 \]

Answer: (A) is propanenitrile, (B) is propiophenone, and (C) is benzoic acid.

Answer 2

Concept: Radioactive decay always follows first-order kinetics. So we use the first-order tools: the rate constant $k = \dfrac{0.693}{t_{1/2}}$ and the time formula $t = \dfrac{2.303}{k}\log\dfrac{N_0}{N}$, where $N_0$ is the starting amount and $N$ is the amount left.

Step 1: Write the rate constant
From the half-life: \[ k = \frac{0.693}{1.5\times10^{10}}\ \text{yr}^{-1} \]

Step 2: Set up the ratio $N_0/N$
The activity drops to 75% of the start, so out of every $100$ we are left with $75$. That gives: \[ \frac{N_0}{N} = \frac{100}{75} = \frac{4}{3} \]

Step 3: Plug into the time formula
Put everything in and notice that $\dfrac{0.693}{k}$ is just $t_{1/2}$: \[ t = \frac{2.303}{k}\log\frac{4}{3} = \frac{2.303}{0.693}\times t_{1/2}\times(\log 4 - \log 3) \]

Step 4: Put in the numbers
Using $\log 4 = 0.60$ and $\log 3 = 0.48$, their difference is $0.12$: \[ t = \frac{2.303}{0.693}\times 1.5\times10^{10}\times(0.60-0.48) = 3.32\times1.5\times10^{10}\times0.12 \] \[ t \approx 5.98\times10^{9}\ \text{years} \]

Answer: The time taken is about $5.98\times10^{9}$ years.

Answer 3

(a) answer: The difference comes from how the two reagents are bonded. KCN is a mostly ionic compound, so in it the cyanide exists as a free $\mathrm{CN^-}$ ion. In this ion the carbon end is the better (more nucleophilic) site, so carbon attacks the haloalkane and we form a C to C bond, giving an alkyl cyanide, $\mathrm{R\text{-}CN}$. AgCN, on the other hand, is mostly covalent, so the carbon lone pair is tied up with silver and is not free; only the nitrogen lone pair is available to attack. So nitrogen bonds to the carbon of the haloalkane, giving an isocyanide, $\mathrm{R\text{-}NC}$.

(b) answer: $\mathrm{S_N2}$ needs the nucleophile to attack the carbon from the back side, so it works best when that carbon is easy to reach and when the C to Cl bond is not too strong. In benzyl chloride the chlorine sits on an $\mathrm{sp^3}$ benzylic carbon, which is open to backside attack, and the bond is a normal single bond, so $\mathrm{S_N2}$ happens easily. In chlorobenzene the chlorine sits directly on an $\mathrm{sp^2}$ aromatic carbon. Here the lone pairs of chlorine go into resonance with the ring, giving the C to Cl bond partial double-bond character, which makes it short and strong, and the flat ring also blocks backside attack. So chlorobenzene is far less reactive towards $\mathrm{S_N2}$ than benzyl chloride.

(c) answer: A Grignard reagent, $\mathrm{RMgX}$, has a very polar carbon to magnesium bond where the carbon is strongly negative, almost like a carbanion. Such a carbon grabs any acidic hydrogen very eagerly, and water has exactly such a hydrogen. So if even a trace of moisture is present, the Grignard reagent is destroyed before it can do its intended job, turning into an alkane: \[ RMgX + H_2O \rightarrow RH + Mg(OH)X \] That is why the reagent must be prepared under dry (anhydrous) conditions.