Multiple Choice QuestionsSN2 mechanism is involed in the following substitution :
CH3-CH2-Cl + OH-
(CH3)2-C(Cl)-CH3-OH-
(CH3)2-C-Cl + OH-
CH3-CH2-(CH3)C(Cl)-CH3 + OH-
A.
CH3-CH2-Cl + OH-
Primary halide undergoes alkaline hydrolysis through SN2 mechanism .
Assertion : The major products formed by heating C6H5CH2OCH, with HI are C6H5CH2I and CH3OH.
Reason : Benzyl cation is more stable than methyl cation.
If both assertion and reason are true and reason is the correct explanation of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion
If assertion is true but reason is false
If both assertion and reason are false.
A.
If both assertion and reason are true and reason is the correct explanation of assertion
This can be explained on the basis of SN1 mechanism, the carbonium ion produced being benzylium ion, since this type is more stable than alkylium ion.
The major product formed in the following reaction is
(CH3)2CH-CH2OCH3
CH3-CH(OCH3)-CH2CH3
CH3-C(CH3)=CH2
(CH3)2-C-OCH3- CH3
D.
(CH3)2-C-OCH3- CH3
In the above reaction, alkyl halide is 1°. It always gives substitution reaction by SN2/E-2 mechanism.
E.g. with (CH3)3C-O-. It gives mainly elimination reaction and involves carbocation intermediate, i.e.primary carbocation. It rearranges itself to form tertiary carbocation. It is mainly because of the stability of a charged system which is increased by dispersal of the charge. Stability of carbocation is 3° > 2° > 1°. Rearrangement can be done in 2 ways-
Therefore,
Which of the following compound gives iodoform test?
CH3CN
CH3OH
CH3COCH3
C6H5OH
C.
CH3COCH3
Compounds containing CH3CO group give the iodoform test. The reaction is
CH3CO-CH3 + 3I2 +NaOH CHI3 + HCOONa + H2O
iodoform (yellow ppt)
Hence, only acetone gives the iodoform test.
Arrange the hydra - acids of halogens in increasing order of acidity .
HF < HCl < HBr < HI
HI < HBr < HCl < HF
HF < HBr < HI <HCl
HF < HI < HBr < HCl
A.
HF < HCl < HBr < HI
Hydra - acids of halogens in increasing order of acidity :
HF < HCl < HBr < HI
As the strength of hydrogen bonding decreases in the order of :
HF > HCl > HBr > HI
IUPAC name of
CH3CH2C(Br)=CH-Cl is :
2 - bromo - l - chloro butene - 1
1 - chloro - 2 - bromo - butene
3 - chloro - 2 - bromo butene - 2
none of these
A.
2 - bromo - l - chloro butene - 1
CH3CH2C(Br)=CH-Cl
2 - bromo - 1 - chloro butene - 1
Assertion : 2-bromobutane on reaction with sodium ethoxide in ethanol gives 1-butene as a major product.
Reason : 1-butene is more stable than 2-butene.
If both assertion and reason are true and reason is the correct explanation of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion
If assertion is true but reason is false
If both assertion and reason are false.
D.
If both assertion and reason are false.
2-bromobutane on reaction with sodium ethoxide in ethanol gives 2-butene as a major product.
This is according to Saytzeffs rule i.e. the predominant product is the most substituted alkene. 2-butene is more stable than 1-butene due to presence of large number of hyperconjugating structures in 2-butene.
Strongest nucleophile is
RNH2
ROH
C6H5O-
CH3O-
D.
CH3O-
CH3O- is stronger nucleophilic reagent because methyl group increase the electron density on oxygen, since it show +I effect.
The major product obtained on treatment of CH3CH2CH(F)CH3 with CH3O/CH3OH is
CH3CH2CH(OCH3)CH3
CH3CH = CHCH3
CH3CH2CH = CH2
CH3CH2CH2CH2OCH3
B.
CH3CH = CHCH3
According to Saytzeff's rule, the major product will be the one that contains more number of substituents around the double bond. Therefore, the correct answer is CH3CH = CHCH3.
Among the following the dissociation constant is highest for
C6H5OH
C6H5CH2OH
CH3CCH
CH3NH3+Cl
D.
CH3NH3+Cl
is a salt which can be completely ionized in aqueous solution.
