Question

Please answer all. Just need A,B,C or D

2. (01') Which proton has the lowest pKa value (This is Elvitegravir an HIV drug)? На CI O2Hd A) Ha B) Hb C) Hc D) Hd 3. (Q1) Which is the major product in this reaction? 1) Sia28H-THF 2) H202, NaOH C) D) но он 4. (a1') What is the hybridization of the nitrogen atom in the following structure? NH A) sp B) sp C) sp D) sp

Answer 1

(2) pKa value is the index to express the acidity of acids. Smaller or lower the pKa value stronger the acid.

We know that the strong acid (which has lower pKa value) have more capacity to donate its proton in the solvent. Here in the given structure (Elvitegravir) has many protons, in which,

• proton Ha is a alcoholic proton (-OH). This has less capacity to donate its proton than carboxylic acid. Its pKa value is high (around 17)
• proton Hb is a ether (-OCH3) proton. This is also has more pKa value because of the weak dissociation capacity.
• proton Hc is attached to the benzene. So it is pKa value is too high (more than 45) because dissociation capacity is too less.
• proton Hd is a carboxylic acid proton (-COOH). This has good dissociation capacity and it donates its proton to the system compared to other proton in this structure. Its pKa value is less (around 4)

Therefore, Hd proton has lowest pKa value in this structure. So, the answer is, D) Hd

(3) This reaction has two steps.

First step: Addition of Sia₂BH to triple bond

The given starting material has triple bond. This will react with Sia₂BH (disiamylborane) and gives a disiamylborane adduct. This reagent helps to the formation of regioselective alkenes.  Here the addition takes place according to the anti-Markovnikov's rule (addition of hydrogen to the more substituted carbon atom).

Second step : Oxidation

In this step oxidation of adduct takes place. Since there is a -BSia₂ group at the less substituted carbon atom of starting material, this group will oxidize to give a terminal aldehyde (cyclopentylacetaldehyde) using H₂O₂ (hydrogen peroxide) with NaOH as a oxidizing agent.

Step 1 CH BSia Sia2BH THF (solv ent) disiamylborane adduct Step 2 H202 NaOH BSia2 cyclopentylacetaldehyde

So, overall if we speak, Hydroboration followed by oxidation of a terminal alkyne gives an aldehyde.

Therefore, the answer is, A) cyclopentylacetaldehyde

4) To find the hybridization of nitrogen atom in the given structure, first we need to understand its bonding with the other atom.

Here nitrogen has a lone pair (it will not involve in the hybridisation) and it is bonded with two carbon atoms (N-C bond) which is a sigma (single) bond and one bond (single) with hydrogen (N-H).

We know that in the outermost shell of Nitrogen has 5 electrons (n=5, p orbital). Among that, 3 electrons are hybridized (one s orbital and two p orbitals are involved in the hybridization).

Thus the hybridization of nitrogen atom in this structure is sp² (there are two carbon atom attached to nitrogen and each overlap with p-orbital of nitrogen and one hydrogen overlap with one s-orbital of nitrogen)

Therefore, the correct answer is , B) sp²