
\(\text{CH}_3(\text{CH}_2)_5\text{COOC}_2\text{H}_5 \xrightarrow{\text{DIBAL-H, H}_2\text{O}} \text{CH}_3(\text{CH}_2)_5\text{CHO}\)
\(\text{C}_6\text{H}_5\text{COC}_6\text{H}_5 \xrightarrow{\text{Zn(Hg) \& conc. HCl}} \text{C}_6\text{H}_5\text{CH}_2\text{C}_6\text{H}_5\)
\(\text{C}_6\text{H}_5 \xrightarrow{\text{CH}_3\text{MgBr, H}_2\text{O}} \text{C}_6\text{H}_5\text{CH(OH)CH}_3\)
\(\text{CH}_3\text{COCH}_2\text{COOC}_2\text{H}_5 \xrightarrow{\text{NaBH}_4, \text{H}^+} \text{CH}_3\text{CH(OH)CH}_2\text{COOC}_2\text{H}_5\)
To solve this problem, we need to match the reactions in List I with the appropriate reagents in List II. Let's analyze each reaction:
Thus, by matching our understanding with our provided options, the correct answer should be:
A - (III), B - (IV), C - (I), D - (II)
An amount of ice of mass \( 10^{-3} \) kg and temperature \( -10^\circ C \) is transformed to vapor of temperature \( 110^\circ C \) by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice = 2100 J kg\(^{-1}\) K\(^{-1}\),
specific heat of water = 4180 J kg\(^{-1}\) K\(^{-1}\),
specific heat of steam = 1920 J kg\(^{-1}\) K\(^{-1}\),
Latent heat of ice = \( 3.35 \times 10^5 \) J kg\(^{-1}\),
Latent heat of steam = \( 2.25 \times 10^6 \) J kg\(^{-1}\))
Match List-I with List-II.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]
Match List-I with List-II.
Choose the correct answer from the options given below :
