General Organic Chemistry MCQ Questions & Answers in Organic Chemistry | Chemistry
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261.
The hybridisation of carbons of $$C - C$$ single bond of $$HC \equiv C - CH = C{H_2}$$ is
A
$$s{p^3} - s{p^3}$$
B
$$sp - s{p^2}$$
C
$$s{p^3} - sp$$
D
$$s{p^2} - s{p^3}$$
Answer :
$$sp - s{p^2}$$
No explanation is given for this question. Let's discuss the answer together.
262.
Arrange the following in decreasing order of solubility in water
A
I > III > II
B
III > II > I
C
II > III > I
D
All are equally soluble
Answer :
I > III > II
Higher the electron density on $$O,$$ stronger is the $$H$$ - bond with water and thus more is the solubility. Thus solubility of the three ethers follow the order
263.
The $$Cl - C - Cl$$ angle in 1, 1, 2, 2 - tetrachloroethene and tetrachloromethane respectively will be about
A
$${120^ \circ }\,\,{\text{and}}\,\,{109.5^ \circ }$$
B
$${90^ \circ }\,\,{\text{and}}\,\,{109.5^ \circ }$$
C
$${109.5^ \circ }\,\,{\text{and}}\,\,{90^ \circ }$$
D
$${109.5^ \circ }\,\,{\text{and}}\,\,{120^ \circ }$$
The bond angle in $$s{p^3},s{p^2}$$ and $$sp$$ hybridization is respectively $$109.28',{120^ \circ }$$ and $${180^ \circ }.$$
Tetrachloroethene being an alkene has $$s{p^2}$$ hybridised $$C $$ - atoms and hence the $$Cl - C - Cl$$ angle is $${120^ \circ },$$ whereas in tetrachloromethane, carbon is $$s{p^3}$$ hybridised, so the angle is $${109^ \circ }.28'.$$
264.
IUPAC name of $${\left( {C{H_3}} \right)_3}C - CH = C{H_2}$$ is
A
2, 2-dimethylbut-3-ene
B
2, 2-dimethylpent -4-ene
C
3, 3-dimethylbut -1-ene
D
hex-1-ene
Answer :
3, 3-dimethylbut -1-ene
3, 3-Dimethylbut -1-ene
265.
The correct order of increasing basicity of the given conjugate bases $$\left( {R = C{H_3}} \right)$$ is
A
$$RCO\overline O < HC \equiv \overline C < \overline R < \overline N {H_2}$$
B
$$\overline R < HC \equiv \overline C < RCO\overline O < \overline N {H_2}$$
C
$$RCO\overline O < \overline N {H_2} < HC \equiv \overline C < \overline R $$
D
$$RCO\overline O < HC \equiv \overline C < \overline N {H_2} < \overline R $$
Answer :
$$RCO\overline O < HC \equiv \overline C < \overline N {H_2} < \overline R $$
The correct order of basicity is $$RCO\overline O < CH \equiv {C^ - } < N{H_2} - < {R^ - }$$
266.
Which one of the following acids would you expect to be the strongest?
A
$$I - C{H_2}COOH$$
B
$$Cl - C{H_2}COOH$$
C
$$Br - C{H_2}COOH$$
D
$$F - C{H_2}COOH$$
Answer :
$$F - C{H_2}COOH$$
Fluorine is most electronegative atom and exerts maximum $$-I$$ effect. Hence, $$F - C{H_2}COOH$$ is the strongest acid.
267.
The pair of electron in the given carbanion, $$C{H_3}C \equiv {C^ - },$$ is present in which orbitals?
A
$$s{p^3}$$
B
$$s{p^2}$$
C
$$sp$$
D
$$2p$$
Answer :
$$sp$$
$$\eqalign{
& {\text{Hybridisation}} \cr
& = \frac{{{\text{Number of }}\sigma {\text{ - }}\,{\text{electrons}}}}{2} \cr
& = \frac{{2 + 2{\text{(negative ion)}}}}{2} \cr
& = 2 \cr
& = sp \cr} $$
Hence, in the carbanion, $$C{H_3}C \equiv {C^\Theta },$$ pair of electron as $$\left( - \right)ve$$ charge is present in $$sp$$ - hybridised - orbital.
268.
Which of the following acid does not exhibit optical isomerism?
A
Maleic acid
B
$$\alpha $$ - amino acid
C
Lactic acid
D
Tartaric acid
Answer :
Maleic acid
Only those compounds exhibit optical isomerism, which have chiral centre and/or absence of symmetrical elements. ( Chiral carbon is the carbon in which all the four valencies are satisfied by four different groups. )
Thus, maleic acid does not exhibit optical isomerism. NOTE
If $$R = H,$$ the $$\alpha $$ - amino acid is achiral.
269.
Two possible stereo-structures of $$C{H_3}CHOH \cdot COOH,$$ which are optically active, are called