An aliphatic hydrocarbon is the fifth member of an unsaturated hydrocarbon series. The compound A rapidly de-colourises Br2/ CCl4.The compound A doesn’t give precipitate with ammoniacal cuprous chloride solution . on complete hydrogenation with H2/Ni, A reacts with two moles of H2 for each mole of A and yields compound B. The B when passed over chrominum oxide at 770 K yields compound C. Compound A on reaction with sodium in the presence of liquid ammonia yields compound D. The compound D can exhibit geometric isomerism.
Q1. The molecular formula of the compound A is : (a)C6H10 (b) C6H12 (c) C5H8 (d) C5H12
Q2. Compound D is : (a) Pent- 1-ene (b)cis-Hex-2-ene (c) trans- Hex-2-ene (d) trans- pent-2-ene
Q3. The geometrical isomer of D on ozonolysis would yield (a) Hexanal (b)Propanal (c)Butanal and ethanal (d)Propanal and propene
Q4. How many isomeric alkynes ( only position isomers)3are possible for the molecular formula A (a) 3 (b) 7 (c) 5 (d) 8
Answers
Answer:
The hydration of propyne with HgSO
4
/H
2
SO
4
followed by keto enol tautomerism gives propan-2-one. Oxidation with concentrated nitric acid gives a mixture of acetic acid and formic acid. Acetic acid reacts with PCl
5
to form acetyl chloride.
Acetyl chloride undergoes esterification with ethanol to form ethyl acetate. Ethyl acetate is also formed when acetic acid reacts with ethanol in presence of conc. H
2
SO
4
.
In Hell Volhard- Zelinsky (HVZ) reaction, acetic acid reacts with Cl
2
/ red P to form monochloro acetic acid. Further chlorination gives dichloro acetic acid and trichloro acetic acid.
Answer:
1) b is the correct answer (C6H12)
2)
3)b is the correct answer (propanal)
4)An alkyne which contains four carbon atoms in the chain has two structural isomers: 1-butyne and 2-butyne. 1-Butyne is a terminal alkyne, with the first three carbon atoms in the chain adopting a linear shape because of the triple bond.