Generate Quiz 6889EAB4BAD42

Since 2 hydroxy propanoic acid is commonly referred to as “lactic acid” and has the formula C3H6O3, Option D is accurate.

The SN reaction occurs when the RO- of alcohol replaces the -OH group of the acid. So, alcohol is the source of bridge oxygen.

is accurate since intermolecular hydrogen bonds determine the boiling point. The presence of carboxylic acid as a cyclic dimer is demonstrated by the molecular mass determination in non-polar solvents such as benzene.

Because of the stronger I-effect of the two alkyl groups and the increased Stearic hindrance provided to the attacking nucleophile, acetone is less reactive to nucleophilic addition reactions than acetaldehyde.

Acetic acid, or CH3COOH, is the weakest acid among the options provided. Of the acids listed, acetic acid has the lowest pKa value at 4.76. A molecule’s acidity is gauged by its pKa value. A molecule is more acidic if its pKa value is lower. The pKa values of the other acids are as follows:

CICH2COOH: pKa = 2.86

Cl2CHCOOH: pKa = 2.78

Cl3CCOOH: pKa = 2.67

As you can observe, acetic acid has a higher pKa value than the other acids. This indicates that acetic acid is less acidic than the other acids. Since acetic acid does not fully ionize in water, it is a weak acid. In a solution, only around 1% of the acetic acid molecules will ionize. The remaining molecules of acetic acid will continue to exist as molecules.

An electrophilic carbon atom is attacked by a nucleophile in an SN2 reaction, which turns the carbon chiral. The stereocenter configuration inverts as a result of the nucleophile attacking from the opposite side of the leaving group. The product has the opposite stereochemistry from the starting material as a result of this configuration inversion. Because the SN2 reaction is stereospecific, the stereochemistry of the product is determined by the stereochemistry of the starting material.

CH3-CH2-OH is an alcohol because it is an organic compound with at least one hydroxyl functional group (-OH) attached to a saturated carbon atom.

The following is a representation of the reaction:
CH3CH2CH2COCH3 + O2 → CH3COOH + CH3CH2CH2COOH
Therefore, butanoic acid and ethanolic acid will be the end products. Therefore, ethanoic acid will be the right response.

Since alkanes are sp3 hybridized and halogens can only be introduced to molecules by substituting (replacing another group), they undergo free radical substitution reactions with halogens.

Three steps are involved in the mechanism of free radical substitution: initiation, propagation, and termination. Diatomic halogen molecules are first broken up into two free radical species, which can subsequently join the alkane molecules during the propagation stages.

The stability of these free radicals rises as their electronegativity falls down the group. The fluorine free radical releases the greatest energy during the propagation stage and attacks the alkane more easily since it is the least stable and most electronegative. The least reactive of them is iodine, which comes after chlorine and bromine. In actuality, it is not reactive enough to remove a H atom from an alkane.

Option C is the right response since this order makes sense.