Calculate Isomers for C3H5Br
To determine the number of isomers for the molecular formula $\text{C}_{3}\text{H}_{5}\text{Br}$, we first calculate the degree of unsaturation (DoU) and then systematically identify all possible structural arrangements.
Determine Degree of Unsaturation (DoU)
The degree of unsaturation formula is:
$ \text{DoU} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1 $For the formula $\text{C}_{3}\text{H}_{5}\text{Br}$ ($C=3, H=5, X=1$):
$ \text{DoU} = 3 - \frac{5}{2} - \frac{1}{2} + 0 + 1 = 3 - 2.5 - 0.5 + 1 = 1 $A DoU of 1 signifies that the molecule contains either one double bond or one ring.
Identify Alkenyl Isomers (One Double Bond)
We examine structures containing a carbon-carbon double bond and one bromine atom.
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3-bromoprop-1-ene: $\text{CH}_{2}=\text{CH}-\text{CH}_{2}\text{Br}$
This isomer does not exhibit geometric (E/Z) isomerism because the first carbon atom of the double bond is attached to two identical hydrogen atoms.
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2-bromoprop-1-ene: $\text{CH}_{2}=\text{CBr}-\text{CH}_{3}$
Similar to the above, this isomer lacks geometric (E/Z) isomerism because the first carbon atom of the double bond has two identical hydrogen atoms.
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1-bromoprop-1-ene: $\text{CH}_{3}-\text{CH}=\text{CHBr}$
This structure shows geometric (E/Z) isomerism. Each carbon atom in the double bond is bonded to two different groups (e.g., $\text{CH}_3$ and H on one, Br and H on the other).
This leads to two distinct isomers:
(E)-1-bromoprop-1-ene
(Z)-1-bromoprop-1-ene
These possibilities yield a total of 4 unique isomers containing a double bond.
Identify Cyclic Isomers (One Ring)
We consider structures featuring a ring and one bromine atom.
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Bromocyclopropane: $\text{C}_{3}\text{H}_{5}\text{Br}$
A cyclopropane ring with a bromine substituent results in only one unique isomer. This is because all carbon atoms in the cyclopropane ring are equivalent due to symmetry, making the position of the bromine atom indistinguishable.
This gives 1 isomer containing a ring structure.
Total Isomer Count
The total number of isomers is the sum of isomers found from each possibility (double bond and ring):
Total Isomers = (Isomers with double bond) + (Isomers with ring)
Total Isomers = 4 + 1 = 5
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Important Questions from Organic Chemistry
- In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ______ %.
(Aqueous tension at 300 K is 15 mm). - $0.25 \text{ g}$ of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $\text{CaCl}_2$ tube and potash tube at the end of the experiment. The amount was found to be $0.15 \text{ g}$ and $0.1837 \text{ g}$, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer)
(Given: molar mass in $\text{g mol}^{-1}$ $\text{H} : 1, \text{C} : 12, \text{O} : 16$) - Grignard reagent $\text{RMgBr (P)}$ reacts with water and forms a gas $\text{(Q)}$. One gram of Q occupies $1.4 \text{ dm}^3$ at STP. (P) on reaction with dry ice in dry ether followed by $\text{H}_3\text{O}^+$ forms a compound $\text{(Z)}$. 0.1 mole of $\text{(Z)}$ will weigh _________ g. (Nearest integer)
- Consider all the structural isomers with molecular formula $\text{C}_5\text{H}_{11}\text{Br}$ are separately treated with $\text{KOH(aq)}$ to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is ___________.
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