Organic Chemistry - Some Basic Principles and Tech

Uncertainty in speed = 90 $*\frac{5}{100}=4.5m/s=\Delta V$ $\frac{}{}$

$\Delta x=\frac{h}{4\pi m\Delta V}=\frac{6.63*10^{-34}}{4*3.14*10^{-2}*4.5}=1.173*10^{-33}$

the nearest integer is 1

Please find the below image

Compounds which are more acidic than phenol

NaCN converted into HCN and Na₂S converted into H₂S

$\begin{array}{l}NaCN+HN{O}_3\to NaN{O}_3+HCN\\ N{a}_{2}S+2HN{O}_3\to NaN{O}_3+H_{2}S\end{array}$

% of C in organic compound

$=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$  

$=\frac{952.56}{21.648}=44\mathrm{\%}$  

$\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$  

= $\frac{20}{18}*\frac{0.4428}{0.492}*100$  

=  $\frac{88.56}{8.856}=10\mathrm{\%}$  

= 100 – 54 = 46%

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

Mass of C =  $\frac{330}{44}*12$ gm = 90 gm

Mass of H =  $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

In metamers, distribution of alkyl groups are changed with respect to polyvalent functional groups.

Structure (I) is anti conformer.

Structure (II) is fully eclipsed conformer.

Structure (III) is skew or gauche conformer.

Structure (IV) is partially eclipsed.

$\therefore$ order of stability

(I) > (III) > (IV) > (II)

Order of P.E. is (II) > (IV) > (III) > (I).

Sodium fusion extract is boiled with conc.   HNO₃ to remove NaCN and Na₂S.

Degree of unsaturation = 4 + 1 -   $\left(\frac{5-1}{2}\right)=3$

sp³ hybridized only one carbon