Q3. Give the organic product of (i) propanoic acid with PCl5, (ii) benzoic acid with ethanol and a little concentrated H2SO4, (iii) ethanoic acid with Br2 and red phosphorus, followed by water.
(i) Propanoyl chloride, CH3CH2COCl, with POCl3 and HCl. (ii) Ethyl benzoate, C6H5COOC2H5, with water. (iii) Bromoethanoic acid, BrCH2COOH, by the Hell–Volhard–Zelinsky reaction (reasoned from NCERT pages 251–253).
Q4. Which reagent would you use to turn 4-nitrobenzoic acid into (4-nitrophenyl)methanol, O2NC6H4CH2OH, without touching the nitro group?
Diborane, B2H6, followed by H3O+: it reduces the carboxyl group to –CH2OH but does not easily reduce a nitro group (NCERT page 253). Ordinary sodium borohydride conditions would not achieve this carboxyl reduction.
Q5. What hydrocarbon does sodium 2-methylpropanoate, (CH3)2CHCOONa, give (a) on heating with soda lime, (b) by Kolbe electrolysis?
(a) Propane, CH3CH2CH3: the –COONa is replaced by H. (b) The expected Kolbe coupling product is 2,3-dimethylbutane, (CH3)2CH–CH(CH3)2: two (CH3)2CH– groups join, twice the three carbons of the alkyl group (reasoned from NCERT page 253). This identifies the coupling product, not a guarantee that a secondary radical gives it exclusively.
Q6. Arrange CH3COOH, FCH2COOH, CHF2COOH and CF3COOH in increasing order of acidity, and give the reason.
CH3COOH < FCH2COOH < CHF2COOH < CF3COOH: each fluorine atom on the α-carbon withdraws electrons and stabilises the carboxylate ion further, as the chlorine atoms do in NCERT’s list (reasoned from NCERT page 251; trifluoroacetic acid has pKa 0.23, page 250).
Q7. 4-Methoxybenzoic acid has pKa 4.46 (NCERT page 251). What is its Ka, and is it a weaker or a stronger acid than benzoic acid?
Ka = 10−4.46 = 3.5 × 10−5. Its pKa is greater than benzoic acid’s 4.19, so it is the weaker acid: the methoxy group donates electrons to the ring (reasoned from NCERT page 251).
Q8. Which of (CH3)2CHCOOH and (CH3)3CCOOH undergoes the Hell–Volhard–Zelinsky reaction, and what does it give with Br2?
(CH3)2CHCOOH, which has one α-hydrogen: it gives 2-bromo-2-methylpropanoic acid, (CH3)2CBrCOOH. (CH3)3CCOOH has no α-hydrogen, so it does not react (reasoned from NCERT page 253).