A benzene derivative did not produce white precipitate with the ammonical silver nitrate solution but decolorised the cold dilute alkaline $\mathrm{KMnO}_4$ solution. The compound is
$\mathrm{C}_8 \mathrm{H}_6$
$\mathrm{C}_8 \mathrm{H}_{10}$
$\mathrm{C}_8 \mathrm{H}_8$
$\mathrm{C}_7 \mathrm{H}_8$
It is a derivative of benzene and decolorised the cold alkaline $\mathrm{KMnO}_4$ solution.Which of the following can be used as the test for unsaturation with regard to colour change of reaction?
Addition of hydrogen
Addition of hydrogen bromide
Addition of hypobromous acid
Addition of bromine
Ethyl phenyl acetylene (1-phenyl-1-butyne) on reduction with partially deactiveated palladised charcoal (Lindlar's catalyst) gives
Which one of the following methods is suitable to generate aromatic compound(s) from linear aliphatic saturated hydrocarbons with at least six carbon atoms?
Heating at 773 K
$\mathrm{Mo}_2 \mathrm{O}_3, 773 \mathrm{~K}, 10-20 \mathrm{~atm}$
Anhyd. $\mathrm{AlCl}_3$, conc. HCl, $\Delta$
$\mathrm{Cu}, 523 \mathrm{~K}, 100 \mathrm{~atm}$
'A' is :
H3C–CH=CH2 $\buildrel {C{l_2}/{H_2}O} \over \longrightarrow $
CH $ \equiv $ CH, CH3–C $ \equiv $ CH and CH2 = CH2 is as follows
Explanation:
Compound (P) has total number of hydroxyl groups = 6
X is:
Among the following reaction(s), which gives(give) tert-butyl benzene as the major product is(are)
In the following reactions, the product S is
The major product U in the following reactions is
In the following reaction, the major product is
The correct statement with respect to product Y is
Among P, Q, R and S, the aromatic compound(s) is(are)
The major product of the following reaction is
The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is _______.
Explanation:
(2) Since, bromide (Br–) is a good leaving group, elimination using strong base takes place using strong base (KOH) take via E2 mechanism. This is also called $\beta $ elimination.
(3) There are 3 different types of protons :
(iv) The strong base abstracts $\beta $ hydrogen (H1 or H2 or H3) with simultaneous loss of bromide ion forming an alkene. This alkene can exist in 2 conformations, E and Z.
(a) Elimination of H1 proton :
This product can exist in 2 conformations, E and Z.
(b) Elimination of H2 proton :
Since, the groups about sp2 hybridised carbon of cyclopentane is fixed no E and Z forms are possible. This molecule has only one conformation.
(c) Elimination of H3 proton :
This product can exist in 2 conformations, E and Z.
Hence, a total of 5 products are possible.
The maximum number of isomers (including stereoisomers) that are possible on mono-chlorination of the following compound, is ____________.
Explanation:
(i) The monochlorination products obtained depends upon the kinds of hydrogen present in the molecule.
(ii) There are four different types of hydrogen in the molecule; $\mathrm{H}_a, \mathrm{H}_b, \mathrm{H}_c$ and $\mathrm{H}_d$.
(iii) The monochlorination products obtained on replacement of these hydrogen are as follows :
(a) Replacing $\mathrm{H}_a$ with chlorine :
Since, the carbon is a chiral; only one monochlorination isomer is possible when $\mathrm{H}_a$ is replaced by chlorine.
(b) Replacing $\mathrm{H}_b$ with chlorine :
Since all the carbons are a chiral; only one monochlorination product is possible by when $\mathrm{H}_b$ replaced by Cl .
(c) Replacing $\mathrm{H}_c$ with chlorine
The carbon on which hydrogen is replaced by chlorine becomes chiral; hence R/S isomers exist.
Also, carbon adjacent to the carbon on which hydrogen $\left(\mathrm{H}_c\right)$ is replaced, is also chiral; hence, it will also exist as a pair of enantiomer $R / S$ four monochlorination products (which are stereoisomers or optical isomers) are possible.
(d) Replacing $\mathrm{H}^d$ by chlorine :
The central carbon $\left({ }^*\right)$ becomes chiral when $\mathrm{H}^d$ is replaced by chlorine; hence $R / S$ isomers exist. Two monochlorination products are possible (stereoisomers or optical isomers) are possible.
Total number of isomers possible are 8.
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ${C_6}{H_5}C{H_2}CH\left( {OH} \right)CH{\left( {C{H_3}} \right)_2}\buildrel {conc.{H_2}S{O_4}} \over \longrightarrow \,?$
