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Chemistry-

General

Easy

Question

Reaction of esters with Grignard reagents give rise to:

primary alcohols
secondary alcohols
tertiary alcohols
ketones

The correct answer is: tertiary alcohols

Related Questions to study

The set of values of x for which the inequality $blank s i n to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses plus c o s to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses less or equal than fraction numerator 1 over denominator 2 end fraction text end text$ holds is

The set of values of x for which the inequality $blank s i n to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses plus c o s to the power of 4 end exponent invisible function application open parentheses fraction numerator x over denominator 3 end fraction close parentheses less or equal than fraction numerator 1 over denominator 2 end fraction text end text$ holds is

The equation |sinx| + |cosx| = 3/2 has

The equation |sinx| + |cosx| = 3/2 has

Among the following compounds which can be dehydrated very easily?

Among the following compounds which can be dehydrated very easily?

Chemistry-General

parallel

$I f blank stack lim with x rightwards arrow infinity below invisible function application open parentheses fraction numerator x plus h over denominator x minus h end fraction close parentheses to the power of x end exponent equals 4 comma blank t h e n blank h equals$

$I f blank stack lim with x rightwards arrow infinity below invisible function application open parentheses fraction numerator x plus h over denominator x minus h end fraction close parentheses to the power of x end exponent equals 4 comma blank t h e n blank h equals$

Arrange the following limits in descending order
i) $stack lim with x rightwards arrow 3 to the power of minus end exponent below invisible function application fraction numerator 3 minus x over denominator open vertical bar x minus 3 close vertical bar end fraction$
ii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator left parenthesis 1 plus x right parenthesis to the power of 1 divided by 8 end exponent minus left parenthesis 1 minus x right parenthesis to the power of 1 divided by 8 end exponent over denominator x end fraction$
iii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator t a n x minus s i n x over denominator x to the power of 3 end exponent end fraction$
iv) $stack lim with x rightwards arrow infinity below invisible function application open curly brackets square root of x to the power of 2 end exponent plus 6 x plus 7 end root minus x close curly brackets$

Arrange the following limits in descending order
i) $stack lim with x rightwards arrow 3 to the power of minus end exponent below invisible function application fraction numerator 3 minus x over denominator open vertical bar x minus 3 close vertical bar end fraction$
ii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator left parenthesis 1 plus x right parenthesis to the power of 1 divided by 8 end exponent minus left parenthesis 1 minus x right parenthesis to the power of 1 divided by 8 end exponent over denominator x end fraction$
iii) $stack lim with x rightwards arrow 0 below invisible function application fraction numerator t a n x minus s i n x over denominator x to the power of 3 end exponent end fraction$
iv) $stack lim with x rightwards arrow infinity below invisible function application open curly brackets square root of x to the power of 2 end exponent plus 6 x plus 7 end root minus x close curly brackets$

$I f blank open square brackets x close square brackets d e n o t e s blank t h e blank g r e a t e s t blank i n t e g e r blank l e s s blank t h a n blank o r blank e q u a l blank t o blank x blank t h e n blank$ $stack lim with n rightwards arrow infinity below invisible function application fraction numerator 1 over denominator n to the power of 4 end exponent end fraction open curly brackets open square brackets 1 to the power of 3 end exponent x close square brackets plus open square brackets 2 to the power of 3 end exponent x close square brackets plus open square brackets 3 to the power of 3 end exponent x close square brackets plus horizontal ellipsis plus left square bracket n to the power of 3 end exponent x right square bracket close curly brackets equals$

$I f blank open square brackets x close square brackets d e n o t e s blank t h e blank g r e a t e s t blank i n t e g e r blank l e s s blank t h a n blank o r blank e q u a l blank t o blank x blank t h e n blank$ $stack lim with n rightwards arrow infinity below invisible function application fraction numerator 1 over denominator n to the power of 4 end exponent end fraction open curly brackets open square brackets 1 to the power of 3 end exponent x close square brackets plus open square brackets 2 to the power of 3 end exponent x close square brackets plus open square brackets 3 to the power of 3 end exponent x close square brackets plus horizontal ellipsis plus left square bracket n to the power of 3 end exponent x right square bracket close curly brackets equals$

parallel

$stack lim with x rightwards arrow 0 below invisible function application open parentheses c o s e c x minus fraction numerator 1 over denominator x end fraction close parentheses equals$

$stack lim with x rightwards arrow 0 below invisible function application open parentheses c o s e c x minus fraction numerator 1 over denominator x end fraction close parentheses equals$

If f (x) = $fraction numerator x over denominator sin invisible function application x end fraction text and g end text left parenthesis x right parenthesis text = end text fraction numerator x over denominator tan invisible function application x end fraction$ where 0 < x ≤ 1 then in the interval

If f (x) = $fraction numerator x over denominator sin invisible function application x end fraction text and g end text left parenthesis x right parenthesis text = end text fraction numerator x over denominator tan invisible function application x end fraction$ where 0 < x ≤ 1 then in the interval

The function 2|x – 2| + |x – 3| is an increasing function when x belongs to the interval

The function 2|x – 2| + |x – 3| is an increasing function when x belongs to the interval

parallel

The function 2x³ – 9x² + 12x – 3 is increasing when x belongs to the interval

The function 2x³ – 9x² + 12x – 3 is increasing when x belongs to the interval

There is an error of 0.05cm in measuring the circumference 20cm of a circle. Then the percentage error in the area is

There is an error of 0.05cm in measuring the circumference 20cm of a circle. Then the percentage error in the area is

In the curve y = a log(x² – a²) the sum of the lengths and the sub tangent at any point varies as

In the curve y = a log(x² – a²) the sum of the lengths and the sub tangent at any point varies as

parallel

If the tangent at any point of the curve $square root of x plus square root of y equals square root of a$ meets the co-ordinate axes at A and B then OA + OB =

If the tangent at any point of the curve $square root of x plus square root of y equals square root of a$ meets the co-ordinate axes at A and B then OA + OB =

The plane 2x -2y + z + 12 = 0 touches the sphere x² + y² + z² – 2x – 4y + 2z – 3 = 0 at the point

The plane 2x -2y + z + 12 = 0 touches the sphere x² + y² + z² – 2x – 4y + 2z – 3 = 0 at the point

$stack lim with x rightwards arrow 0 below invisible function application fraction numerator open parentheses a b close parentheses to the power of x end exponent minus a to the power of x end exponent minus b to the power of x end exponent plus 1 over denominator x to the power of 2 end exponent end fraction equals$

$stack lim with x rightwards arrow 0 below invisible function application fraction numerator open parentheses a b close parentheses to the power of x end exponent minus a to the power of x end exponent minus b to the power of x end exponent plus 1 over denominator x to the power of 2 end exponent end fraction equals$

parallel

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