Q2.In the following reaction, the compound used in the reaction for synthesizing ethyl fluoride is – +AgF → H3C – F + AgBr
Answer : Option CExplaination / Solution:
Alkyl
fluorides is only formed by Swartz reaction.in swartz reaction alkyl
chlorides or bromide reacts with metallic flouride agf hg2f2 to form alkyl flourides.
Q3.Which compound in the following pair reacts faster in SN2 reaction with OH– ?
i) CH3Br or CH3
ii) CH3Cl, (CH3)3CCl
Answer : Option AExplaination / Solution:
For SN2 reaction leaving group should be good and alkyl halide should be primary.so CH3Br and CH3Cl are 10organic compounds hence rate of sn2 reaction will be more.
Answer : Option DExplaination / Solution:
For SN1 stability of carbocation which matters. In C6H5C(CH3)(C6H5)Br carbocation will be tertiary and will be resonance stablised.
Q5.Upon oxidation, one of the following halogen containing compounds forms an extremely poisonous gas – – + O2 → 2 COCl2 + 2HCl
Answer : Option CExplaination / Solution:
CHCl3 on reaction with oxygen forms phosgene. Chloroform
decomposes when it exposed to sunlight.it decomposes to a harmful gas
phosgene(COCl2) Chloroform decomposes when it exposed to sunlight.it decomposes to a harmful gas phosgene(COCl2)
Q6.Carbon – halogen bond of alkyl halides is responsible for their nucleophilic substitution, elimination and their reaction with metal atoms to form organometallic compounds because of their
Answer : Option DExplaination / Solution:
R—X is polar.
Q7.Which one of the following is not a chiral molecule?
Answer : Option CExplaination / Solution:
.this
is not chiral molecule because functional group carbon that is carbon
having alchohal group has 2 same groups on either side and there is
symmentry about this carbon that is why this is not chiral.
Q8.A mixture containing two enantiomers in equal proportions
Answer : Option DExplaination / Solution:
For a mixture of two enantiomers , one rotates the plane polarized light towards left and other towards right therefore there is no optical rotation and hence mixture is optically inactive.
Q9.
In the reaction Pent – 2 – ene. 2 – Bromopentane. Pent – 1 – ene 2 –
Bromopentane on heating with alcoholic KOH, forms two compounds: Pent – 1
– ene and Pent – 2 – ene. Which one of the following statements is true
Answer : Option DExplaination / Solution:
Pent-2-ene is major product as ore substituted alkene is formed in major quantity according to zaitsev rule.
Pent – 2 – ene. 2 – Bromopentane. Pent – 1 – ene,
2 – Bromopentane on heating with alcoholic KOH, forms two compounds:
Pent – 1 – ene and Pent – 2 – ene., if one major and one product are
formed , then
Answer : Option BExplaination / Solution:
The major alkene will ne more substituted that what zaitsev rule is.
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0