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Question: 10. $CF_3I + NaOH \longrightarrow A+B$ $CH_3I + NaOH \longrightarrow P+Q$ (i) pcc (ii) $CH_3MgBr$ ...

  1. CF3I+NaOH⟶A+BCF_3I + NaOH \longrightarrow A+B

CH3I+NaOH⟶P+QCH_3I + NaOH \longrightarrow P+Q (i) pcc

(ii) CH3MgBrCH_3MgBr

(iii) H3O⊕H_3O^{\oplus}

(iv) B

C+DC+D (sodium salt of poisonous acid)

Find the correct statement.

A

A>CA > C (Boiling point)

B

A<CA < C (Vapour pressure)

C

A<CA < C (Water solubility)

D

A>CA > C (Dipole moment)

Answer

A < C (Water solubility)

Explanation

Solution

Explanation of the Solution:

  1. The reaction of CF₃I with NaOH gives a product A that is essentially a fluorocarbon (fluoroform, HCF₃) which is very volatile and hardly soluble in water.
  2. In the sequence starting from CH₃I + NaOH, methanol (P) is generated and then oxidized (by PCC) to give formaldehyde. Its subsequent reaction with CH₃MgBr and work‐up, followed by the reaction with “B” (the other product from CF₃I reaction) produces an oxygenated product C (an acylated/alcohol derivative) whose structure contains polar functional groups. In addition, D is the sodium salt of trifluoroacetic acid (a very poisonous acid).
  3. Because product C bears polar (and hydrogen‐bonding) groups it will be more water–soluble than the non–polar compound A.
  4. Hence, we have A < C in water solubility.