Introduction To Organic Chemistry

To determine the empirical formula of the organic compound, we start by understanding that empirical formulas represent the simplest whole-number ratio of atoms in a compound. Given:

• The compound contains carbon by weight.
• The remaining percentage is hydrogen, which is (since ).
• The atomic weights are: and .

Next, calculate the moles of each element in 100 grams of the compound:

For Carbon:

For Hydrogen:

Next, determine the simplest whole-number ratio of carbon to hydrogen by dividing both by the smallest number of moles:

Moles of C:

Moles of H: , which is approximately 3 when rounded to the nearest whole number.

The simplest whole-number ratio of carbon to hydrogen is . Therefore, the empirical formula of the compound is .

Therefore, the correct option is: .

Let's analyze the given statements regarding chirality to identify the incorrect one:

1. SN₁ reaction yields a 1:1 mixture of both enantiomers. This statement is correct. The SN₁ reaction involves a carbocation intermediate that is achiral, allowing the nucleophile to attack from either side, resulting in a racemic mixture.

2. The product obtained by SN₂ reaction of haloalkane with chirality at the reactive site shows inversion of configuration. This statement is correct. SN₂ reactions involve a backside attack which leads to inversion of configuration at the chiral center.

3. Enantiomers are superimposable mirror images on each other. This statement is incorrect. Enantiomers are non-superimposable mirror images of each other; their mirror images do not align perfectly when superimposed.

4. A racemic mixture shows zero optical rotation. This statement is correct. A racemic mixture contains equal amounts of two enantiomers, and their optical rotation cancels each other out, resulting in no net optical rotation.

Based on the analysis, the incorrect statement is: Enantiomers are superimposable mirror images on each other.

Pyridine has a chemical formula of C₅H₅N. The structure of pyridine consists of a ring of six carbon atoms and one nitrogen atom. Each carbon atom is bonded to two other carbon atoms and one hydrogen atom, and the nitrogen atom is bonded to one carbon atom and two hydrogen atoms. The nitrogen atom also possesses a lone pair of electrons. The aromatic ring has alternating single (σ) and double (π) bonds. 

Counting the bonds and lone pairs:

• Carbon-carbon (C-C) single bonds: 5 (in the ring)
• Carbon-hydrogen (C-H) single bonds: 5 (one on each carbon)
• Carbon-nitrogen (C-N) single bond: 1
• Pi (π) bonds: 3 (these are part of the aromatic system)
• Lone pair of electrons on nitrogen: 1

Adding these up gives:

• Sigma (σ) bonds: 5 + 5 + 1 = 11
• Pi (π) bonds: 3
• Lone pair of electrons: 1

So, the correct option is (B): 11, 3, 1

The given question discusses the identification test involving Lassaigne's extract, which is utilized to detect the presence of nitrogen and sulfur in organic compounds. When both elements are present, a specific colored reaction occurs.

The Lassaigne’s extract is first reacted to form sodium thiocyanate (NaSCN). In the presence of ferric ions (Fe³⁺), NaSCN forms a complex with the ferric ions. The specific complex responsible for the blood red coloration is the thiocyanatoiron complex, given by the chemical formula [Fe(SCN)]²⁺.

This complex occurs because the thiocyanate ion (SCN^-) acts as a ligand, binding to the ferric ion to form the colored complex. The reaction can be summarised as:

Thus, the correct answer, based on the given options, is:

To determine which compound is most readily dehydrated under acidic conditions, we need to consider the stability of the carbocation intermediate formed during the dehydration process. Dehydration under acidic conditions typically involves the following steps:

1. Protonation of the hydroxyl group to form water, a better leaving group.
2. Loss of water to form a carbocation intermediate.
3. Rearrangement if necessary to form a more stable carbocation.
4. Elimination of a proton to form a double bond, resulting in an alkene.

Among the options provided, the compound that forms the most stable carbocation will dehydrate most readily. The stability of carbocations follows the order: tertiary > secondary > primary, due to hyperconjugation and inductive effects.

Let's examine the given options:

Option 3, which forms a tertiary carbocation, is the most stable. Therefore, it will be dehydrated most readily under acidic conditions.