The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is ________.
C4H5N
DBE=(C+1)-(\(\frac{H+X-N}{2}\))
=(4+1)-(\(\frac{5-1}{2}\))=5-2=3
Three double-bond equivalents are present in the compound
Only 1 sp3 hybridised carbon is there
(Keeping compound as acyclic)



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:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
Organic Chemistry is a subset of chemistry dealing with compounds of carbon. Therefore, we can say that Organic chemistry is the chemistry of carbon compounds and is 200-225 years old. Carbon forms bond with itself to form long chains of hydrocarbons, e.g.CH4, methane and CH3-CH3 ethane. Carbon has the ability to form carbon-carbon bonds quite elaborately. Polymers like polyethylene is a linear chain where hundreds of CH2 are linked together.
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Organic chemistry is applicable in a variety of areas including-