Haloalkanes and Haloarenes
Consider all the structural isomers with molecular formula $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Br}$ are separately treated with $\mathrm{KOH}(\mathrm{aq})$ to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is $\_\_\_\_$。
Explanation:
As per the language given and considering the condition we are going with answer 3 and considering both active isomers we will be giving 6 too.
RMgI when treated with ice cold water liberated a gas which occupied $1.4 \mathrm{dm}^3 / \mathrm{g}$ at STP. The gas produced is further reacted with iodine in presence of $\mathrm{HIO}_3$ to give compound $(\mathrm{X})$. Compound $(\mathrm{X})$ in presence of Na and dry ether produced compound $(\mathrm{Y})$. Molar mass of compound $(\mathrm{Y})$ is $\_\_\_\_$ $\mathrm{g} \mathrm{mol}^{-1}$. (Nearest integer)
Explanation:
Given that the gas produced by treating RMgI with ice-cold water occupies a volume of $1.4 \, \mathrm{dm}^3$ per gram at STP.
At STP, one mole of any gas occupies $22.4 \, \mathrm{dm}^3$. So, the molar mass of the gas can be calculated as:
$ \text{Molar mass} = \frac{22.4}{1.4} \approx 16 $
This shows that the gas produced is methane, $\mathrm{CH_4}$.
The methane formed reacts with iodine in the presence of $\mathrm{HIO_3}$ to form methyl iodide ($\mathrm{CH_3I}$), which is compound $(\mathrm{X})$.
When $\mathrm{CH_3I}$ is treated with sodium metal in dry ether, a coupling reaction (Wurtz reaction) occurs, forming ethane ($\mathrm{C_2H_6}$), which is compound $(\mathrm{Y})$.
The molar mass of ethane is:
$ \text{Molar mass of } \mathrm{C_2H_6} = (2 \times 12) + (6 \times 1) = 30 \, \mathrm{g \, mol^{-1}} $
Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is $80 \%$ whereas Q is $100 \%$.
Mass of product Q obtained is________ g.
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{O}: 16, \mathrm{Br}: 80$ )
Explanation:
$\begin{aligned} & \text { Molecular mass of } \mathrm{Q}=230 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \begin{aligned} \text { Mass of } \mathrm{Q} & =0.8 \times 230 \\ & =184 \mathrm{~g} \end{aligned} \end{aligned}$
Consider the following sequence of reactions:
11.25 mg of chlorobenzene will produce ___________$\times 10^{-1} \mathrm{mg}$ of product B.
(Consider the reactions result in complete conversion.)
[Given molar mass of $\mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{N}$ and Cl as $12,1,16,14$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively]
Explanation:
$\begin{aligned} & \frac{11.25 \times 10^{-3}}{112.5}=\frac{x \times 10^{-1} \times 10^{-3}}{93} \\ & x \times 10^{-1}=93 \times 0.1 \\ & x=93 \mathrm{mg} \end{aligned}$
The ratio of number of oxygen atoms to bromine atoms in the product Q is _________ $\times 10^{-1}$.
Explanation:
Number of O-atoms $=3$
Number of $\mathrm{Br}$-atoms $=2$
$\begin{aligned} \text { Required ratio } & =\frac{3}{2}=1.5 \\ & =15 \times 10^{-1} \end{aligned}$
The number of halobenzenes from the following that can be prepared by Sandmeyer's reaction is _________
Explanation:
Only II and III can be prepared from Sandmeyer’s reaction.
The total number of hydrogen atoms in product A and product B is _________.
Explanation:
Total number of hydrogen atom in A and B is 10
2-chlorobutane $+\mathrm{Cl}_2 \rightarrow \mathrm{C}_4 \mathrm{H}_8 \mathrm{Cl}_2$ (isomers)
Total number of optically active isomers shown by $\mathrm{C}_4 \mathrm{H}_8 \mathrm{Cl}_2$, obtained in the above reaction is _________.
Explanation:
Number of moles of AgCl formed in the following reaction is _____________
Explanation:
Circled Cl will get precipitated
In the presence of sunlight, benzene reacts with Cl2 to give product, X. The number of hydrogens in X is _____________.
Explanation:
Total number of hydrogens are 6.
The major product of the following reaction contains ____________ bromine atom(s).
Explanation:
Here only carbon 12th and 13th is not taking part in the resonance. All other carbons are taking part in resonance and lone pair of O is also taking part in the resonance.
As because of resonance stability of compound increases Br will not attack to any carbon which is taking part in the resonance and decrease the stability of compound.
So, Br can only attack 12th carbon or 13th carbon atom but 12th carbon is a alpha carbon and it can easily donate H+ ion and add Br- ion and increase stability.
Consider the above reaction. The number of $\pi$ electrons present in the product 'P' is __________.
Explanation:
The given reaction undergoes nucleophilic substitution by SN2 mechanism at room temperature
$\therefore $ No. of $\pi$ electrons present in $\mathrm{P}=2$
(i) forms aldehydes on ozonolysis followed by the hydrolysis.
(ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP.
The number of carbon atoms in a molecule of compound A is ___________.
Explanation:
22400 ml of A $\Rightarrow$ ${{1.53} \over {448}}$ $\times$ 22400 gm A = 76.5 Agm
$\mathop {{H_3}CHC - CH - Cl}\limits_{It\,has\,3\,carbon\,atoms} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Zn/{H_2}O}^{{O_3}}} \mathop {C{H_3} - CH = O}\limits_{Aldehyde} $
& MM is 36 + 5 + 35.5 = 76.5
Give reasons :
Explanation:
(A) (i) When (2-bromopropan-2-yl)benzene on reaction with aqueous ethanol, an ether is formed by replacement of bromide ion by ethoxy ion. The products formed are an ether and an acid, hydrogen bromide.
The presence of HBr makes the solution acidic.
(ii) The 1-bromo-4-(propan-2-yl)benzene does not react with aqueous ethanol. Thus, solution is neutral in nature.
(B) (i) The reaction between given compound with aqueous NaOH is an example of bimolecular reaction. The fluoride group at ortho position is replaced with hydroxy group. The fluoride ion is liberated during the reaction. The rate of reaction is enhanced by the presence of electron withdrawing groups at ortho and para positions.
(ii) In this case, a bimolecular reaction does not occur. Thus, fluoride ion does not get eliminated from the compound.
(C) (i) The nitrosobenzene on reaction with mixture of conc. HNO$_3$ and conc. H$_2$SO$_4$ undergoes nitration. The nitrogen atom has a lone pair of electrons on it. Also, nitro so group is both ortho and para directing. Therefore, H atoms at ortho and para will be replaced by nitro group producing mixture of ortho and para products.
(ii) The nitrobenzene on further nitration using the mixture of conc. HNO$_3$ and conc. H$_2$SO$_4$, produces dinitrobenzene, by substituting H atom at meta position by nitro group. The nitro group is electron withdrawing group and is meta directing in nature.
(D) The reaction of compound with hydrogen gas in presence of Pd/C as a catalyst, the compound undergoes reduction producing following product.
In the given compound, there is a central benzene ring attached to three four membered rings which further attached to terminal benzene rings. The three 4 membered rings are antiaromatic in nature. During a chemical reaction, the rings will be undergoing reaction such that most stable product will be formed.
When a central benzene ring will undergo reduction, the stability of three antiaromatic rings will be achieved. Thus, compound obtained is quite stable.
If the reduction of one of the terminal benzene rings occurs, only one four membered antiaromatic ring will attain stability. Thus, the compound produced will be less stable as compared to that of first compound.
Final Answer :
(A) (i) Due to formation of acid, HBr
(ii) No reaction
(B) (i) Bimolecular reaction occurs.
(ii) No bimolecular reaction.
(C) (i) Nitro group is both ortho and para directing in this case and N atom of NO group has a lone pair of electrons on it.
(ii) In this case, nitro group is meta directing.
(D) The reduction of central benzene ring will give stability to 3 four membered antiaromatic rings, producing a stable compound.
Hints :
The reaction between (2-bromopropan-2-yl) benzene with aqueous ethanol, produces an ether with HBr which is an acid. Hence, the solution becomes acidic in nature.