Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 64

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following products will be obtained when neopentyl alcohol is treated with conc. $$HCl$$ in presence of $$Zn{Cl}_{2}?$$

A
t-Butyl chloride

B
Isobutylene

C
t-Pentyl chloride

D
Neo pentyl chloride

Solution

When neopentyl alcohol is treated with conc. $$HCl$$ in presence of $$ZnCl_2,$$ t-pentyl chloride will be final product.
In this reaction, less stable primary carbocation rearranges to more stable tert carbocation through methyl shift. This gives rearranged product.
The overall reaction is the replacement of $$-OH$$ group with chloride.
Question 2
1 / -0

Which of the following order is correct for the decreasing reactivity of ring monobromination of the following compounds?

A
$$I > II > III > IV$$

B
$$II > III > IV > I$$

C
$$I > III > II > IV$$

D
$$III > I > II > IV$$

Solution

Ring monobromination is electrophilic aromatic substitution reaction. Higher is the electron density on aromatic ring, higher will be the reactivity towards ring monobromination. Alkyl substituents increases the electron density by $$+I$$ effect. Hence, the reactivity of toluene (I) is maximum.

Substituents such as $$-COOH$$ group and $$-NO_2$$ group are electron withdrawing groups. hence, they decrease the electron denstiy of the ring.

Hence, nitrobenzene has lowers reactivity towards ring monobromination.

Hence, the correct order for the decreasing reactivity of ring monobromination is $$I$$ (toluene) >$$ III$$ (benzene) > $$II$$ (benzoic acid)> $$IV$$ (nitrobenzene)
Question 3
1 / -0

What will be the product of the following the reaction
$$\underset{|}{CH_3}$$
$$CH_3-C-CH_2-Br\xrightarrow[]{NaOH} P$$,
$$\overset{|}{CH_3}$$

P is ________.

A
$$CH_3-C\! \! \! =CH-CH_3$$
$$\; \; \; \; \; \; \; \; \; \overset{|}{CH_3}$$

B
$$CH_3-CH-CH=CH_2$$
$$\; \; \; \; \; \; \; \; \; \overset{|}{CH_3}$$

C
Both of these

D
None of these
Question 4
1 / -0

$$\xrightarrow{1eqv.\,

of Br_2 / Fe}$$ A. Compound A is:

A

B

C

D

Solution

The position of Bromine depends on substituting group not on the reactant As the $$C$$ atom acquires positive charge due to electronegative chlorine atoms. Hence, the group becomes deactivating and meta-directing, that is it will have negative inductive effect on the ring more than the positive mesomerio effect.
Correct option: $$A$$
Question 5
1 / -0

Directions For Questions

There are number of organic compound including ester, xanthate, amine oxide etc. that undergo pyrolyticelimination with heat in absence of added reagent either in inert solvent or in absence of solvent. Ingeneral these elimination follow the rate low.
Rate $$\mu$$ [substrate]
However reaction is different from $$E_1$$ elimination by the degree of syn stereoselectivity that they exhibit.These reaction are also known as $$E_i$$ elimination and the degree of syn stereoselectivity reflect the extent to which they proceed via cyclic transition state.

...view full instructions

Find the major product of the following reaction.

A

B

C

D
none

Solution

The compound undergoes pyrolytic elimination with heat. The pyrolysis of $$N,N-dimethyl-2-phenylcyclohexylamine-N-oxide$$ shows how conformational effects and the stability of the transition state affect product composition for cyclic substrates.

In the trans isomer, there are two $$cis-\beta-hydrogens$$ that can be eliminated. The major product is the alkene that is in conjugation with the phenyl ring, presumably due to the stabilizing effect on the transition state.

In the cis isomer, there is only one $$cis-\beta-hydrogen$$ that can be eliminated, giving the nonconjugated regioisomer as the major product.
Question 6
1 / -0

An organic compound $$A(C_4H_9Cl)$$ on reaction with $$Na/diethyl \hspace{1mm}ether$$ gives a hydrocarbon which on monochlorination gives only one chloro derivative. $$A$$ is:

A
t-butyl chloride

B
sec-butyl chloride

C
isobutyl chloride

D
n-butyl chloride

Solution

Only single chloro derivative is obtained when the compound is symmetrical.
Question 7
1 / -0

A mixture of two organic compounds was treated with sodium metal in ether solution. Isobutane was obtained as a product. The two chlorine compounds are:

A
methyl chloride and propyl chloride

B
methyl chloride and ethyl chloride

C
isopropyl chloride and Methyl chloride

D
isopropyl chloride and ethyl chloride

Solution

Wurtz reaction:
$$RX + 2 Na + RX \xrightarrow[ether]{pure \, \& \, dry} R - R + 2 NaX$$
$$CH_3Cl + 2 Na + Cl \, CH_3 \xrightarrow[ether]{pure \, \& \, dry} CH_3CH_3 + 2NaCl$$
$$2CH_3 - \underset{CH_3}{\underset{|}{CH}} - Cl + 2Na \xrightarrow[ether]{pure \, \& \, dry} CH_3 \underset{CH_3}{\underset{|}{CH}} - \underset{CH_3}{\underset{|}{CH}} CH_3 + 2NaCl$$
$$RX + 2 Na + XR' \xrightarrow[ether]{pure \, \& \, dry} $$ Mixture of $$R' - R' + R' - R + R - R$$ (mixture of three)
$$CH_3 Cl + 2Na + Cl \, C_2H_5 \xrightarrow[ether]{pure \, \& \, dry} CH_3CH_3 + C_3H_8 + C_4H_{10}$$
Question 8
1 / -0

$$R-\overset { \, \, R_{ 1 } }{ \overset { | }{ \underset { \, R_{ 2 } }{ \underset { | }{ C } } } } -X\xrightarrow [ ]{ HOH } R-\overset { \,\,R_{ 1 } }{ \overset { | }{ \underset { R_{ 2 } }{ \underset { | }{ C } } } } -OH$$
For the given reaction, Which substance will give maximum racemisation?

A
$$H_5C_6 - \!\!\!\overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}{C}}}} \!\!- Br$$

B
$$H_2C = CH - \!\!\!\overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}{C}}}} \!\!- Br$$

C

D

Solution

Maximum racemisation will occur from where there is stable carbocation formed.As in $$C$$ carbocation is highly resonance stabilised.
Question 9
1 / -0

Directions For Questions

There are number of organic compound including ester, xanthate, amine oxide etc. that undergo pyrolyticelimination with heat in absence of added reagent either in inert solvent or in absence of solvent. Ingeneral these elimination follow the rate low.
Rate $$\mu$$ [substrate]
However reaction is different from $$E_1$$ elimination by the degree of syn stereoselectivity that they exhibit.These reaction are also known as $$E_i$$ elimination and the degree of syn stereoselectivity reflect the extent to which they proceed via cyclic transition state.

...view full instructions

Product formed in the reaction is__________.

A

B

C

D

Solution

The pyrolytic decomposition of esters is an example of a thermal syn elimination. Esters containing $$\beta-hydrogens$$ can eliminate a carboxylic acid through a 6-membered transition state, resulting in an alkene.
Question 10
1 / -0

For the following reaction which of the following statement is/are correct?

A
Both $$X$$ and $$Y$$ are five membered cyclic ring

B
Reaction (i) follows $$E2$$ path where as reaction (ii) follows $$E1$$ path

C
Reaction (ii) has carbocation as an intermediate

D
Carbocation rearrangemnet takes place in reaction (i)

Solution

(A) $$X$$ is five membered cyclic ring and $$Y$$ is four membered cyclic ring.
(B) reaction (i) follows $$E1$$ path where as reaction (ii) follows $$E2$$ path.
(C) reaction (ii) does not have a carbocation as an intermediate.
(D) carbocation rearrangement takes place in reaction (i)
Due to 1,2 alkyl shift, 4 member ring expands to 5 member ring. This relieves ring strain.
Hence, only statement (D) is correct.