NEET Practice Questions (MCQs) with Answers & Solutions

An optically active compound that rotates plane-polarized light counterclockwise is known as levorotatory, denoted by '(-)'. Therefore, all the given options (levorotatory, (-), and all of the above) are correct. The correct answer is 'all of above'.

The anomeric carbon in the cyclic structure of glucose is the carbon that was the carbonyl carbon (C=O) in the linear form of the molecule. For glucose, this is carbon 1 ( ext{C}_1). When glucose cyclizes to form a ring (either as an α or β anomer), ext{C}_1 becomes the anomeric carbon.

The anomeric carbon in the cyclic structure of fructose is the carbon that was the carbonyl carbon (C=O) in the linear form of the molecule. For fructose, this is carbon 2 ( ext{C}_2). When fructose cyclizes to form a ring (either as an α or β anomer), ext{C}_2 becomes the anomeric carbon.

The specific rotation of an aqueous solution of sucrose before hydrolysis is ext{+66.5}^ ext{°}. Specific rotation is a characteristic property of a chiral substance and is used to identify specific substances and concentrations.

The phenomenon when the rotation of sucrose solution is reversed is known as 'inversion'. Inversion occurs when sucrose, which is dextrorotatory, hydrolyzes into glucose and fructose, which results in a mixture that is levorotatory. This change in the direction of optical rotation from right to left is called inversion.

Before hydrolysis, sucrose is dextrorotatory, which means it rotates the plane of polarized light to the right. After hydrolysis, the products are glucose and fructose, which result in a levorotatory mixture because fructose has a stronger levorotatory effect compared to the dextrorotatory effect of glucose. Therefore, the optical rotation changes from dextrorotatory to levorotatory.

Sucrose is commonly chosen as the standard for sweetness of sugars. In sweetness comparison, sucrose is assigned a value of 1 (or 100 in some scales), and the sweetness of other sugars is compared relative to sucrose.

When sucrose is heated at 486 K (approximately 213°C), it undergoes a decomposition reaction and forms caramel. This process is known as caramelization, a non-enzymatic browning reaction that produces various compounds responsible for the characteristic color and flavor of caramel.

In maltose, the two glucose units are linked by an α(1→4) glycosidic bond. This means that the first carbon (C1) of one glucose molecule is linked to the fourth carbon (C4) of the second glucose molecule through an oxygen atom. Therefore, the correct option is $ C_1 - O - C_4 $.